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If $M$ is a smooth manifold and $X(t)$ is a parameterized curve on it, then at each point $X(t)$, we can define a tangent vector that lies in $T_{X(t)}M$. All these tangent vectors together form a tangent (to that curve) vector field $v_X$. Why this is called the "velocity" is intuitively clear to me.

But I'm really struggling to intuitively understand why we define acceleration as $\nabla_{v_X}v_X$. The only way I can reconcile this acceleration definition is as follows:

In general, if $Y,Z$ are two vector fields, then should I interpret $\nabla_YZ$ as the change in $Z$ as I move along the curve $\gamma$ to which $Y$ is tangent (i.e. at each point $p$, $Y(p)$ is the tangent vector to $\gamma(p)$)?

Is the above interpretation correct? Should I then say about $\nabla_{v_X}v_X$ that if $X(t)$ is the trajectory of a particle, then $\nabla_{v_X}v_X$ is the change in $v_X$ as I move along that trajectory? Would appreciate any help!

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You are correct in the interpretation. I will try adding a few comments. If you compare to usual vector calculus in flat space, when one has a curve $\vec{r}(t)$, the acceleration corresponds exactly to the second derivate of the trajectory or alternatively the time derivative of the velocity, here all being a curve of vectors in 3D (trajectory, velocity, acceleration)

In space-time trajectory is now a curve of 4-vectors and its derivative with respect to its parameter (which is not neccessarily $t$) gives you immediately tangent vectors at every point. The problem begins with the second derivative in curved space-times, because if you were to naively take the time derivative of the velocity you might end up with objects that don't belong to the manifold (this is easier to depict in 2-dim while traveling on the surface of a sphere). Since we are beings "living" on the manifold itself, the acceleration we observe must be a vector in our manifold (on the sphere in the 2-dim example), so what one does is project this naive time derivative onto the manifold, this is exactly the covariant derivative.

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  • $\begingroup$ Thanks for the great answer! So your remark on naive time derivative of velocity not belonging to the manifold, and of projecting it on the manifold - this is from the point of view of extrinsic differential geometry, right? (i.e. manifold embedded in some higher dim space). $\endgroup$
    – Shirish
    Commented Sep 2, 2020 at 14:04
  • $\begingroup$ Exactly, ideally we don't want to embed (need to) the manifold in an another space. $\endgroup$
    – ohneVal
    Commented Sep 2, 2020 at 14:21

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